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2 years ago

Math problem its the screen shot

Mathematics

1 answer:

satela [25.4K]2 years ago

4 0

Answer:

It would be 18 feet long for each flag.

Step-by-step explanation:

Due to there being a ratio of 3 feet long to 4 feet wide, this would be the ratio:

3 : 4

Now, if it is now 24 feet wide, multiply bothe sides by 6 to get this amount. This is because 4 * 6 = 24 and that the problem involves a ratio.

3 : 4

x6 Both Sides

18 : 24

<em>The answer would be 18 feet long.</em>

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If a point is equidistant from the endpoint of segment, then it is

Mashutka [201]

.... then it is located on the perpendicular bisector.

7 0

3 years ago

The sphere is _____ cubic centimeters bigger than the cube. (Round to the nearest cubic centimeter.)

Misha Larkins [42]

ANSWER

The sphere is 10762 cubic centimeters bigger than the cube.

EXPLANATION

We want to find the difference in the volumes of the sphere and the cube.

To do this, we have to find the volumes of the sphere and cube and subtract that of the cube from the sphere.

The volume of a sphere is given as:

$V=\frac{4}{3}\pi r^3$

where r = radius

The radius of the sphere is 15 centimeters. Therefore, the volume of the sphere is:

$\begin{gathered} V=\frac{4}{3}\cdot\pi\cdot15^3 \\ V\approx14137\operatorname{cm}^3 \end{gathered}$

The volume of a cube is given as:

$V=s^3$

where s = length of the side

The length of the side of the cube is 15 centimeters. Therefore, the volume of the cube is:

$\begin{gathered} V=15^3 \\ V=3375\operatorname{cm}^3 \end{gathered}$

Therefore, the difference in the volumes of the sphere and cube is:

$\begin{gathered} V_d=V_s-V_c \\ V_d=14137-3375 \\ V_d=10762\operatorname{cm}^3 \end{gathered}$

Therefore, the sphere is 10762 cubic centimeters bigger than the cube.

6 0

1 year ago

In an arithmetic sequence, = 5 and = 11.

Free_Kalibri [48]

Answer:

1134.62 (THE SUM OF 20 TERMS)

Step-by-step explanation:

First we write out our given information:

a8=2a2

a11=18

an is an arithmetic sequence.

Where here an means the nth term of our sequence.

What does an arithmetic sequence mean? It means to get to the next term in your sequence you add a constant (c) each time:

an+1=an+c

Equivalently:

an+1−an(n+1)−n=c

So an is of slope c (c2 is another constant):

an=cn+c2

Where here c2=a0 (Substitute in n=0 and see why that has to be the case if we let a0 exist)

Now we use the other given information to try to come up with a solution.

Let n=2:

a2=2c+c2

Let n=8, using the above equation we have:

a8=8c+c2=2a2=4c+2c2

Let n=11

a11=18

a11=11c+c2

But a11−a8=(11c+c2)−(8c+c2)=3c

Hence, a11=3c+a8

a11=3c+4c+2c2=18

a11=3c+8c+c2=18

5 0

3 years ago

Can you help me with 7 and 8

elena55 [62]

All I know or all I understood was both 7 and 8 aren’t functions. Hope this helped

4 0

3 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var

-BARSIC- [3]

Answer:

A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}$

Mean = 10.65

Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.9

11.2

11.5

11.8

n=10(even)

$Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7$

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}$

Standard deviation : $\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}$

Standard deviation :0.89916

8.3

,10.3

,10.5

,10.6

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

,10.3

,10.5

,10.6

,10.7

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5$

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2$

IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range : $(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)$

Range :(9.45, 10.15)

8.3 does not lie in this interval

So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

3 0

3 years ago